We have a signal $$x(t)=U_1\cos\omega_1t+U_2\cos\omega_2t$$ whose frequencies are $f_1=100\,\text{Hz}$ and $f_2=600\,\text{Hz}$. This signal is being discretisated in time using unipolar rectangular pulse with 1 V amplitude, length $\tau$ and period $T=\frac{1}{f_0}$. Can we reconstruct this signal without distortion if we use ideal low-pass filter if sampling frequency is:

a) $f_0=1\,\text{kHz}$
b) $f_0=1.3\,\text{kHz}$

My attempt:

when we want to tell whether we can or cannot reconstruct original signal when sampling with specific frequency we compare sampling frequency with the frequency of the given signal, sampling frequency must be at least as twice big as signal frequency (NYQUIST criterion). However i cannot determine frequency of this signal since it has two components.

Since we are doing time discretisation with rectangular pulse we want to find fourier transform of $f(t)=x(t)s(t)$ where $s(t)=\sum_{n=-\infty}^{\infty} \frac{\tau}{T}sinc(n\omega_0\tau/2)e^{jn\omega_0t}$.

I know that multiplication in time is convolution in frequency but i don't know how to implement it in this case since $X(j\omega)=\frac{U_1}{2}2\pi\delta(\omega -\omega_1)+\frac{U_1}{2}2\pi\delta(\omega +\omega_1) + \frac{U_2}{2}2\pi\delta(\omega -\omega_2) + \frac{U_2}{2}2\pi\delta(\omega +\omega_2)$

How can i do that and how can i tell, when i find fourier transform, if signal can be reconstructed or not? Any help appreciated!

  • 1
    $\begingroup$ small corrections: 1. Nyquist demands more than twice as high and 2. Nyquist demands more than twice as high as the highest-frequency component. So, the fact that your signal is a sum of two signals doesn't make any difference. $\endgroup$ Commented May 30, 2018 at 18:00
  • 1
    $\begingroup$ In other words, you don't need "the frequency of the signal" but its bandwidth, defined as the highest frequency component present in the signal. $\endgroup$
    – MBaz
    Commented May 30, 2018 at 18:05
  • $\begingroup$ I see, frequency i am looking for here is 600Hz, which now makes the answer obvious, anyway, i am asked to confirm what i concluded by nyqusit criterion by drawing a graph of signal i got in frequency domain, all i got is a bunch of dirac impulses at frequencies 100, 400, 600... hertz and i don't see much of a difference there whether i use first or second low-pass filter so i am a bit confused about proving that this indeed is true by doing some math (that was, btw, what i was actually asked to do). $\endgroup$
    – cdummie
    Commented May 30, 2018 at 19:01

1 Answer 1


The question is trying to make you understand Aliasing

The low-pass filter is introduced (as a hint) to imply that the output is band-limited to half the Sampling frequency (Why ? See below for aliases).
This would allow you to understand how out of band signals can sneak into your sampled signals if left unchecked. Look at p5 here

You should read in detail about sampling. Skipping the real math, as rule of thumb you can do as below to answer your questions quick

a) for $f_0=1\,\text{kHz}$ the signal cannot be reconstructed

Once you sample your signal with a rate you create aliases at frequencies
$\pm f_{signal}\pm n*f_{sampling}$
where n can take any positive value

If you calculate all combinations for all your frequencies you end up with
[100 400 600 900 1100 1600 1900 2100 2900 ...]

Since you'd sampled with 1000Hz, and are using an (ideal) LPF at 500Hz, you'd take all components from the set above between [0-500Hz] in which case you end up with [100 400].

Do you see the 400Hz odd man out here?
This is termed as the alias of the 600Hz signal.

So, for an input of [100 600] and you end up with [100 400].

b) for $f_0=1.3\,\text{kHz}$ reconstruction works

Repeat steps above.

You'll find that the only components between [0-650Hz] are [100 600].

There you go !

  • $\begingroup$ I see, i finally understand this better! Thanks, I appreciate it! $\endgroup$
    – cdummie
    Commented Jun 2, 2018 at 14:16

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